In Young's double slit experiment intensity at a point is `((1)/(4))` of the maximum intersity. Angular position of this point is

Options

A. `sin^(-1) ((lamda)/(d))`
B. `sin^(-1) ((lamda)/(2d))`
C. `sin^(-1) ((lamda)/(3d))`
D. `sin^(-1) ((lamda)/(4d))`

Answer

C

Explanation

`I= I_("max") cos^(2) ((phi)/(2)) (I_("max"))/(4)= I_("max") "cos"^(2) (phi)/(2)`
`"cos"(phi)/(2)= (1)/(2)`
`(phi)/(2)= (pi)/(3)`
`rArr phi= (2pi)/(3) = ((2pi)/(lamda))xx Delta x` …(i)
where, `Delta x= d sin theta`
Putting the value of `Delta x` in Eq. (i), we get
`sin theta = (lamda)/(3d)`
`theta= sin^(-1) ((lamda)/(3d))`